Question Number 164237 by DAVONG last updated on 15/Jan/22
Answered by mathmax by abdo last updated on 15/Jan/22
$$=\mathrm{lim}_{\mathrm{x}\rightarrow−\mathrm{1}} \:\:\:\:\frac{\mathrm{nx}^{\mathrm{n}−\mathrm{1}} −\mathrm{n}}{\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)}\:=\mathrm{lim}_{\mathrm{x}\rightarrow−\mathrm{1}} \frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)\mathrm{x}^{\mathrm{n}−\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\left(\mathrm{hospital}\:\mathrm{2times}\right) \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 15/Jan/22
$$\mathrm{l}=\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}}\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{2}} =−\frac{\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)}{\mathrm{2}} \\ $$